Question

find the value of x and y (i) root 2 +root 3 /3 root 2 -2 root 3 = x +root 6y​

Answer 1

(root 3 + root 2)(root 3 - root 2)
(a+b)(a-b) = a square - b square
(root 3 + root 2)(root 3 - root 2) = {root 3} ^2 - {root 2 }^2
= 3- 2
(root 3 + root 2)(root 3 - root 2) = 1

Answer 2

$\sf{\displaystyle\frac{\sqrt{2}+\sqrt{3}}{3 \cdot \sqrt{2}-2 \cdot \sqrt{3}}=x+\sqrt{6 y}}$

$\to\sf{\displaystyle x+\sqrt{6}y=\frac{\sqrt{2}+\sqrt{3}}{3\sqrt{2}-2\sqrt{3}}}$

$\to\sf{\displaystyle x+\sqrt{6}y-\sqrt{6}y=\frac{\sqrt{2}+\sqrt{3}}{3\sqrt{2}-2\sqrt{3}}-\sqrt{6}y}$

$\to\sf{\displaystyle x=\frac{12+5\sqrt{6}}{6}-\sqrt{6}y}$